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MySQL Cookbook by Paul DuBois

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Using a Join to Fill in Holes in a List

Problem

You want to produce a summary for each of several categories, but some of the categories are not represented in the data to be summarized. Consequently, the summary has missing categories.

Solution

Create a reference table that lists each category and produce the summary based on a LEFT JOIN between the list and the table containing your data. Then every category in the reference table will appear in the result, even “empty” ones.

Discussion

When you run a summary query, normally it produces entries only for the values that are actually present in the data. Let’s say you want to produce a time-of-day summary for the records in the mail table, which looks like this:

mysql> SELECT * FROM mail;
+---------------------+---------+---------+---------+---------+---------+
| t                   | srcuser | srchost | dstuser | dsthost | size    |
+---------------------+---------+---------+---------+---------+---------+
| 2001-05-11 10:15:08 | barb    | saturn  | tricia  | mars    |   58274 |
| 2001-05-12 12:48:13 | tricia  | mars    | gene    | venus   |  194925 |
| 2001-05-12 15:02:49 | phil    | mars    | phil    | saturn  |    1048 |
| 2001-05-13 13:59:18 | barb    | saturn  | tricia  | venus   |     271 |
| 2001-05-14 09:31:37 | gene    | venus   | barb    | mars    |    2291 |
| 2001-05-14 11:52:17 | phil    | mars    | tricia  | saturn  |    5781 |
...

To determine how many messages were sent for each hour of the day, use the following query:

mysql> SELECT HOUR(t) AS hour, COUNT(HOUR(t)) AS count
    -> FROM mail GROUP BY hour; +------+-------+ ...

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